Solve for $x$ : $2x^2 - 32x + 120 = 0$
Explanation: Dividing both sides by $2$ gives: $ x^2 {-16}x + {60} = 0 $ The coefficient on the $x$ term is $-16$ and the constant term is $60$ , so we need to find two numbers that add up to $-16$ and multiply to $60$ The two numbers $-10$ and $-6$ satisfy both conditions: $ {-10} + {-6} = {-16} $ $ {-10} \times {-6} = {60} $ $(x {-10}) (x {-6}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -10) (x -6) = 0$ $x - 10 = 0$ or $x - 6 = 0$ Thus, $x = 10$ and $x = 6$ are the solutions.